Apologies for the very simple question, but I can't seem to find a reference one way or the other, and it's been bugging me for a while now.

Is there a compact (Hausdorff, or even T1) (topological) group which is infinite, but has countable cardinality? The "obvious" choices don't work; for instance, $\mathbb{Q}/\mathbb{Z}$ (with the obvious induced topology) is non-compact, and I get the impression that profinite groups are all uncountable (although I might be wrong there). So does someone have an example, or a reference in the case that there are no such groups?

8 Answers
8

No, there is no countably infinite compact Hausdorff topological group.

Indeed such a group $G$ would have a left-invariant Haar measure $m$ with $m(G)=1$
and all points would have the same measure (since the group acts transitively on itself).
But then, by countable additivity of the measure $m$, the group itself would have measure $m(G)=0$ or $m(G)=\infty$ according as its points all had $m(p)=0$ or $m(p)>0$ . A contradiction in both cases to the fact that $m(G)=1$ .

A Baire category argument shows that any countable, locally compact Hausdorff group must be discrete. Of course, to moreover be compact it would have to be finite.

In more detail (apologies if this is known/tedious): in any locally compact Hausdorff topological space, the intersection of a countable collection of open dense subsets is dense - the proof is basically the one usually taught for complete metric spaces, I don't know off-hand where to find the LCH case but Kelley's book seems an obvious first guess. From this you can show that any countable locally compact Hausdorff top space must contain an isolated point (open and closed). [Cf. Konstantin's answer.] Since we're in a topological group, translations are homeomorphisms and thus every point is isolated, i.e. the space is discrete.

Munkres' book "Topology" (in section 48 of the second edition) proves that any compact Hausdorff space is a Baire space, and gives the case of locally compact Hausdorff spaces as an exercise.
–
Ricardo AndradeJul 9 '13 at 7:48

@YemonChoi How can one translate this question in term of Hopf algebras?
–
Ali TaghaviJul 19 '14 at 17:36

1

@AliTaghavi I don't know, and if you are seeking a quantum group version then there is the problem of deciding what should replace the notion of a "point", and what one means by a "countable quantum group". Possibly you could try to claim that a quantum group G is "countable" if $L^1({\bf G})$ is separable, and then show that this is incompatible with $C_0({\bf G})$ being unital (which is one way of saying that the quantum group ${\bf G}$ is compact). If you are not familiar with the quantum group formalism then I suggest looking at various expository papers on arXiv by van Daele
–
Yemon ChoiJul 19 '14 at 20:17

EDIT: Both Georges's and Yemon's arguments are better since they avoid explanation why the group has to be metrizable.

No, there is no countably infinite compact group. The reason is that such group would be metrizable and hence a compact Polish space without isolated points. Into any such space one can embed the Cantor set, which is uncountable. The latter is not hard to prove, or you can look at Kechris, Classical Descriptive Set Theory, Theorem 6.2.

The identity component of a compact Hausdorff group is a connected, normal Hausdorff space, so by Urysohn's lemma, if it has two points it is uncountable. Therefore a countable compact Hausdorff group is totally disconnected. But then it is a profinite group, and it must be finite, for infinite profinite groups are uncountable.

Given a compact Hausdorff space without isolated points, one can easily construct an embedding of the Cantor set into it: take two points, separate them with closure-disjoint open sets, repeat iteratively in each balls' closure. Since we started with a group, it is homogeneous, so either no points are isolated - and hence there is an embedded Cantor set of cardinality $2^\omega$ - or all points are isolated, which implies finiteness due to compactness.

I like the idea, but is it really that easy to show that the construction gets you a big enough set? In a metric space I agree that this intersection will contain a copy of the Cantor set, but in a general LCH space I seem to be missing something. (I'm not saying the construction doesn't work, but that it seems to require a bit more slogging than your answer suggests.)
–
Yemon ChoiDec 8 '09 at 17:30

Right, this may not end up with a set homeomorphic to the Cantor set, but rather one with a surjection onto a Cantor set. Let $A^{s0}_n$ and $A^{s1}_n$ ($s$ is a length $n-1$ bit string) be the two disjoint closed sets within $A^{s}_{n-1}$, with the starting set $A^{\emptyset}_0$, the whole space. For each chain $A^{b_1}_1 \supset A^{b_1b_2}_2 \supset ...$ of closed (hence compact) sets their intersection will be nonempty; in general, it may not be a single point, but mapping it to the corresponding point $\sum \frac{b_j}{3^j}$ in the Cantor set will give you a surjective (and continuous) map.
–
ThornyDec 9 '09 at 9:54

It is a well known fact (see for example Hodel´s article on the Handbook of Set Theoretic Topology) that the cardinality of an infinite compact homogeneous space is always a power of $2$. Topological groups are homogeneous (as it is pointed out in Yemon´s answer) and therefore any infinite compact group has the size of the continuum or bigger.